Products Remember, Sums Forget
Another logic puzzle for this week. Four tiny Latin squares share a cross of cells, and each corner hides one bent cage: two tell you what the cells multiply to, two what they add to — and one kind quietly tells you far more. Three checks, then the board resets. Can you fill all twenty-five?
A 3×3 Latin square is the smallest Sudoku there is: the digits 1, 2 and 3, each used exactly once in every row and every column. On its own it is no puzzle at all — hand someone two cells and they will fill the other seven without thinking. So we won’t give you one. We’ll give you four, and make them share.
Below is a 5×5 board. Hidden in it are four 3×3 squares — top-left, top-right, bottom-left and bottom-right — outlined in teal. Each square shares a row or a column with each of its neighbours — the top pair share a column, the left pair share a row, and so on — and all four meet at the center cell. Each of the four squares must be a Latin square in its own right: 1, 2, 3 once in every row of that square and once in every column. A digit in the shared cross has to keep two squares happy at once. The center cell has to keep four.
Then there are the cages — four of them, one tucked into each corner, each bending around a corner of its own. A cage marked × tells you what its three cells multiply to; a cage marked + tells you what they add to. The two product cages live in the top-left and bottom-right squares, the two sum cages in the other pair. One more rule, and it matters: a cage may repeat a digit, so long as the repeats never share a row or a column of a square. For a cage that turns a corner, that means the two ends are allowed to match. The middle cell may match neither.
Tap a cell to cycle it through 1, 2, 3 and blank, or click it and type. When every cell is filled, press Check. You get three checks. Each one tells you how many cells are breaking a rule — not which ones — and after the third miss the board locks until you reset it. So the game isn’t to guess quickly. It’s to be certain before you ask.
one square two squares all four cage
Tap a cell to cycle 1 → 2 → 3 → blank. Arrow keys move; 1, 2, 3 and Backspace work too.
Two of the four clues are a good deal more generous than the other two. Working out which — and what to do about the stingy pair — is most of the puzzle. When you’re done, the solution explains why the generous ones had to be generous, and why that fact reaches all the way down to the foundations of arithmetic.
Explore this puzzle visually with an interactive diagram — drag sliders, watch the geometry update in real time, and build intuition before you solve.
A cage that turns a corner can hold the same digit twice. Its two end cells sit in different rows and different columns of their square, so nothing forbids them from matching — and in a three-cell cage, the ends are the only cells that can. Now write 12 as a product of three digits from 1, 2, 3, where the two ends may match but the middle may match neither. There is exactly one way. Then do the same with a sum of 5, and count how many ways you find.
The finished board, read row by row: 3 2 1 3 2 / 2 1 3 2 1 / 1 3 2 1 3 / 2 1 3 2 1 / 3 2 1 3 2. The center is a 2. The two product cages hand over their cells on sight; the two sum cages don’t — and the reason they don’t is the point of the puzzle.
Step 1 — the product cage gives itself away
Three cells, product 12, digits 1 to 3 only. Because the cage bends, its two ends may match, but the middle cell — the corner of the L — must differ from both. Now list every way to make 12: 2·2·3 is the only one (3·4·1 needs a 4; 1·1·12 needs a 12). Since only the ends may repeat, the two 2s are the ends and the 3 is the middle. The top-left corner of the board is a 3, flanked by a 2 to its right and a 2 below it.
Now watch those 2s. One sits in row 1 and one in row 2; one in column 1 and one in column 2 — all inside the top-left square. That square’s third 2 has nowhere left to go but row 3, column 3: the center of the whole board is a 2. The rest of the square falls by elimination. Row 1 needs its 1 at column 3. Column 3 now holds a 1 and a 2, so row 2 ends in a 3, which leaves a 1 for the cell beside it. Row 3 reads 1, 3, 2. Nine cells from one clue.
Step 2 — the sum cage hedges
The top-right cage adds to 5. Three digits from 1, 2, 3, with only the ends allowed to match: 1+3+1 works, and so does 2+1+2. Two different cages, one total. The clue alone cannot decide — and this is where the overlap earns its keep. The cage’s ends sit in rows 1 and 2 of the top-right square, and row 1 of that square already contains a 1: the shared cell at column 3, which you filled in Step 1. So the ends can’t be 1. They’re 2, and the middle is 1. Then row 1 wants its 3 at column 4, column 4 wants its 1 at row 3, and the last cell, row 3 column 5, is a 3.
Notice where the repeated digit landed: on 2, the same as the center. That isn’t luck. A repeated digit in a corner cage occupies two rows and two columns of its square, so its third copy is forced into the one row and the one column it hasn’t touched — the shared ones, which meet at the center. Every corner cage that repeats a digit repeats the center digit. If you spotted that, the whole board was yours the moment the ×12 cage told you the center was a 2.
Step 3 — same trick, other side
The bottom-left cage adds to 7: either 2+3+2 or 3+1+3. Its ends sit in columns 1 and 2 of the bottom-left square, and column 2 already holds a 3 in the shared row. So the ends are 2 (the center digit again, as promised) and the middle, row 5 column 1, is 3. Column 1 now reads 1, 2, 3 top to bottom; column 2 needs its 1 at row 4; row 4 then needs its 3 at column 3; and row 5 closes with a 1.
Step 4 — nothing left to decide
Look at the bottom-right square before you touch it. Its top row (2, 1, 3) came from Step 2 and its left column (2, 3, 1) from Step 3 — and a 3×3 Latin square is pinned completely by one row and one column. Row 4 has a 3 at the shared cell and column 4 already holds a 1, so row 4 finishes 2, 1; column 4 then wants a 3 at the bottom, and the final cell is a 2. The ×6 cage reads 1·2·3. It was never a clue. It was a receipt.
Why products remember and sums forget
Here is every way to fill a bent three-cell cage with 1s, 2s and 3s, given that the ends may match and the middle may match neither:
| Ends · middle | Sum | Product |
|---|---|---|
| 1 · 2 | 4 | 2 |
| 1 · 3 | 5 | 3 |
| 2 · 1 | 5 | 4 |
| 1, 2, 3 in any order | 6 | 6 |
| 2 · 3 | 7 | 12 |
| 3 · 1 | 7 | 9 |
| 3 · 2 | 8 | 18 |
Seven fills. The sums land on only five totals — 5 and 7 are each hit twice — while the seven products are seven different numbers. A sum cage marked 5 or 7 is bluffing; a product cage never can. The reason is the oldest theorem in number theory, the fundamental theorem of arithmetic: every whole number has exactly one factorization into primes. So 12 is 2·2·3 and nothing else, and 9 is 3·3 and nothing else. A product is a complete record of what went into it. A sum is a compression — once 2+3+2 and 3+1+3 have both become 7, the information about which was which is simply gone. Products remember. Sums forget.
One honest caveat, and the board illustrated it: products remember what, not where. The ×6 cage told you its three cells were 1, 2 and 3 but said nothing about their order, which is why it was the weakest clue on the board — and why the puzzle was built so you would never need it.
The hidden stripes
There is a second reason four small cages could pin down twenty-five cells. A 3×3 Latin square on 1, 2, 3 is always a set of diagonal stripes: each row is the row above it shifted one step left, or one step right, and there are no other kinds. That makes exactly 12 of them — six orders for the top row, two directions. Overlap four and the count collapses further. Fix the top-left square and each neighbour has just two choices: continue the stripes across the shared line, or reflect them. The fourth square then has no choice at all. Forty-eight boards in total, and that is why the finished grid is its own mirror image top to bottom: the stripes bounced off the shared row. Across the shared column they ran straight through, which is why the top row reads 3, 2, 1, 3, 2 and keeps going.
Where this shows up
Latin squares were a working tool long before they were a pastime. In the 1920s R. A. Fisher laid out crop trials as Latin squares — each fertilizer once in every row and every column of the field — so that a gradient in the soil could not masquerade as a difference between treatments. The same arrangement schedules round-robin tournaments and exam timetables, and it is the skeleton of Sudoku, a 9×9 Latin square with boxes added. Samurai Sudoku, with five overlapping grids, is this puzzle’s big sibling.
The product–sum asymmetry shows up wherever information has to survive a trip through arithmetic. Gödel numbering packs an entire sequence of symbols into a single product of prime powers, precisely because that product can be unpacked again. Public-key cryptography lives on the other face of the same coin: multiply two enormous primes and the product still remembers them completely — it just takes longer than the age of the universe to make it say so. Sums are what you reach for when forgetting is the goal: a store total that doesn’t itemize, an average that hides the outliers, a checksum that only has to notice that something changed. Knowing which operation you are looking at tells you how much of the past you can expect to recover from it.
One email, one puzzle, no noise — with a hint ladder and a full worked solution.